



3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, 123, 129, 135, 141, 147, 153, 159, 165, 171, 177, 183, 189, 195, 201, 207, 213, 219, 225, 231, 237, 243, 249, 255, 261, 267, 273, 279, 285, 291, 297, 303, 309, 315, 321, 327
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OFFSET

0,1


COMMENTS

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0(37).
Continued fraction expansion of tanh(1/3).
If a 2set Y and a 3set Z are disjoint subsets of an nset X then a(n4) is the number of 3subsets of X intersecting both Y and Z.  Milan Janjic, Sep 08 2007
Leaves of the Odd CollatzTree: a(n) has no odd predecessors in all '3x+1' trajectories where it occurs: A139391(2*k+1) <> a(n) for all k; A082286(n)=A006370(a(n)).  Reinhard Zumkeller, Apr 17 2008
From Loren Pearson, Jul 02 2009: (Start)
Values of n in 2^n1 that produce a composite with 7 as a factor.
Their distribution in 2^n1 sequence equidistant between terms that have multiple factors of 3 (n=6,12,18,24,30,36,... where the number of factors of 3 equals [number of times 3 divides n] + 1), recognizing that all even n in the 2^n1 sequence have at least one factor of 3. Other odd n appear to be unrelated prime or semiprime composites. (End)
Let random variable X have a uniform distribution on the interval [0,c] where c is a positive constant. Then, for positive integer n, the coefficient of determination between X and X^n is (6n+3)/(n+2)^2, that is, A016945(n)/A000290(n+2). Note that the result is independent of c. For the derivation of this result, see the link in the Links section below.  Dennis P. Walsh, Aug 20 2013
Positions of 3 in A020639.  Zak Seidov, Apr 29 2015
a(n+2) gives the sum of 6 consecutive terms of A004442 starting with A004442(n).  Wesley Ivan Hurt, Apr 08 2016
Numbers k such that Fibonacci(k) mod 4 = 2.  Bruno Berselli, Oct 17 2017
Also numbers k such that t^k == 1 (mod 7), where t is a member of A047389.  Bruno Berselli, Dec 28 2017


LINKS

Table of n, a(n) for n=0..54.
Milan Janjic, Two Enumerative Functions
Friedrich L. Bauer, Der (ungerade) CollatzBaum, Informatik Spektrum 31 (Springer, April 2008).
Tanya Khovanova, Recursive Sequences
Luis Manuel Rivera, Integer sequences and kcommuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
William A. Stein, Dimensions of the spaces S_k^{new}(Gamma_0(N))
William A. Stein, The modular forms database
Dennis P. Walsh, The correlation for a power curve on nonnegative support
Eric Weisstein's World of Mathematics, Collatz Problem
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (2,1).


FORMULA

a(n) = 3*(2*n + 1) = 3*A005408(n), odd multiples of 3.
A008615(a(n)) = n.  Reinhard Zumkeller, Feb 27 2008
A157176(a(n)) = A103333(n+1).  Reinhard Zumkeller, Feb 24 2009
a(n) = 12*n  a(n1) for n>0, a(0)=3.  Vincenzo Librandi, Nov 20 2010
G.f.: 3*(1+x)/(1x)^2.  Mario C. Enriquez, Dec 14 2016


MAPLE

seq(6*n+3, n=0..40); # Dennis P. Walsh, Aug 20 2013
A016945:=n>6*n+3; # Wesley Ivan Hurt, Sep 29 2013


MATHEMATICA

Range[3, 500, 6] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
Table[6 n + 3, {n, 0, 100}] (* Wesley Ivan Hurt, Sep 29 2013 *)
LinearRecurrence[{2, 1}, {3, 9}, 55] (* Ray Chandler, Jul 17 2015 *)
CoefficientList[Series[3 (x + 1)/(x  1)^2, {x, 0, 60}], x] (* Robert G. Wilson v, Dec 14 2016 *)


PROG

From Wesley Ivan Hurt, Sep 29 2013: (Start)
(Haskell)
a016945 = (+ 3) . (* 6)
a016945_list = [3, 9 ..]
(MAGMA) [6*n+3 : n in [0..100]];
(Maxima) makelist(6*n+3, n, 0, 100);
(PARI) {a(n) = 6*n + 3}
(End)
(PARI) x='x+O('x^99); Vec(3*(1+x)/(1x)^2) \\ Altug Alkan, Apr 08 2016


CROSSREFS

Third row of A092260.
Cf. A008588, A016921, A016933, A016957, A016969.
Subsequence of A061641; complement of A047263; bisection of A047241.
Cf. A000225.  Loren Pearson, Jul 02 2009
Cf. A020639.  Zak Seidov, Apr 29 2015
Cf. A004442.
Sequence in context: A030594 A032676 A228935 * A222640 A110108 A162843
Adjacent sequences: A016942 A016943 A016944 * A016946 A016947 A016948


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


STATUS

approved



